Calculate hydraulic radius Rh = A/P and hydraulic diameter Dh = 4A/P for circular, rectangular, trapezoidal, and triangular cross-sections.
The hydraulic radius is a fundamental geometric property used throughout fluid mechanics. Defined as the ratio of cross-sectional flow area to wetted perimeter (Rh = A/P), it characterizes how efficiently a channel shape conveys flow. The closely related hydraulic diameter Dh = 4A/P = 4Rh is used to substitute non-circular cross-sections into pipe-flow equations like Darcy–Weisbach and the Reynolds number.
For a full circular pipe, Rh = D/4 and Dh = D, which is why the hydraulic diameter collapses to the actual diameter. For non-circular ducts — rectangular, trapezoidal, annular — Dh provides the equivalent pipe diameter for friction and heat-transfer calculations. In open channels, the wetted perimeter excludes the free surface, so only the submerged walls count.
This calculator supports five cross-section types: circular (with partial fill), rectangular, trapezoidal, triangular, and custom (direct A and P entry). The fill-level feature for circular pipes is especially useful for sanitary sewer and stormwater design, where pipes rarely run full.
Hydraulic radius and diameter are needed for every pipe, duct, and channel calculation — from pressure-drop estimates in rectangular HVAC ducts to stormwater design in trapezoidal ditches. This calculator handles all common shapes and partial-fill conditions. The note above highlights common interpretation risks for this workflow. Use this guidance when comparing outputs across similar calculators. Keep this check aligned with your reporting standard.
Hydraulic Radius: Rh = A / P Hydraulic Diameter: Dh = 4A / P = 4 Rh Full circular pipe: Rh = D/4, Dh = D Rectangular (open top): P = 2h + w, A = wh Trapezoidal: A = (b + T)/2 × h, P = b + 2√(h² + ((T−b)/2)²)
Result: Rh = 0.075 m, Dh = 0.3 m
A = π/4 × 0.3² = 0.07069 m². P = π × 0.3 = 0.9425 m. Rh = 0.07069 / 0.9425 = 0.075 m. Dh = 4 × 0.075 = 0.3 m (equals the actual diameter).
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The factor of 4 makes Dh equal to the actual diameter for a full circular pipe: D = 4 × (πD²/4)/(πD) = D. This convention simplifies substitution into standard pipe-flow equations.
No. The wetted perimeter includes only surfaces in contact with the fluid. In an open channel, the top (air–water interface) is excluded.
Interestingly, a circular pipe achieves its maximum Rh at about 81% fill — not at 100%. This is because the wetted perimeter grows faster than the area near the top of the circle.
For non-circular cross-sections, Re = ρVDh/μ, where Dh replaces the pipe diameter. This gives the same friction factor from the Moody chart (approximately).
For a given area, a semicircle has the least wetted perimeter and thus the highest Rh. Among practical shapes, a wide-and-shallow trapezoidal channel is more efficient than a deep-and-narrow one.
Yes — Manning's formula v = (1/n) Rh^(2/3) S^(1/2) uses the hydraulic radius directly. It is the standard method for open-channel flow calculations.